The width of a finite potential well

In summary, the conversation discusses approximating the width of a potential well in which an electron is enclosed. By using the time-independent Schrödinger equation and solving for the width of the well, the approximate width is calculated to be 0.73nm.
  • #1
TheSodesa
224
7

Homework Statement


An electron is enclosed in a potential well, whose walls are ##V_0 = 8.0eV## high. If the energy of the ground state is ##E = 0.50eV##, approximate the width of the well.

Answer: ##0.72nm##

Homework Equations


For an electron in a potential well, whose energy is less than the height of the potential walls, outside the well the time-independent Schrödinger equation becomes:
\begin{equation}
\Psi ''(x) = \frac{2m}{\hbar^2}(V_0 - E) \Psi (x) = \alpha^2 \Psi (x),
\end{equation}
where
\begin{equation}
\alpha^2 = \frac{2m}{\hbar} (V_0 - E) > 0
\end{equation}
Inside the well, where ##V(x) = 0##, it is the familiar:
\begin{equation}
\Psi ''(x) = -k^2 \Psi (x), k^2 = \frac{2mE}{\hbar^2}
\end{equation}
By doing a whole bunch of math (by requiring that ##\Psi## be continuously differentiable at the potential walls and solving the resulting linear system with the assumption that it has more than one solution, meaning ##det() = 0##), we and up with the result
\begin{equation}
\frac{\sin (kL)}{\cos (kL)} = \tan (kL) = \frac{\alpha}{k}.
\end{equation}
Substituting ##\alpha## from ##(2)## and ##k## from ##(3)##, we get
\begin{equation}
\tan (kL) = \sqrt{\frac{V_0 - E}{E}}
\end{equation}

The Attempt at a Solution


Now my idea was to use ##(5)## to solve for ##L## as follows:
[tex]
\tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\
\iff\\
kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\
\iff\\
L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}})

= \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\

= \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\
\arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\

= 2.084 751 \cdot 10^{-8}m,
[/tex]
which is a bit off. Any idea what I'm doing wrong?
 
Last edited:
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  • #3
I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

Also, make sure your calculator is in the radian mode for taking the arctan.
 
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  • #4
TSny said:
I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

Also, make sure your calculator is in the radian mode for taking the arctan.

Yeah, it was the radian thing, and the fact that the well goes from -L to L. The online material for our book (Tipler, Modern Physics, itself glosses over this part, the mathematics in particular) actually uses the letter ##a## in place of ##L##, but I figured I could just change the place of the origin without doing anything to make up for it, so to speak.

Thanks again.
 
Last edited:
  • #6
TheSodesa said:
Apparently there are even and odd solutions. The online material for our book states, that the ##-\cot(ka)## actually corresponds to the odd case.
Yes?
 
  • #7
Simon Bridge said:
Yes?

I as just noting that the link (the part of the page) you provided used ##-\cot## instead of ##\tan##. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
 
  • #8
TheSodesa said:
I as just noting that the link (the part of the page) you provided used ##-\cot## instead of ##\tan##. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
Yes? Where are you up to in your solution?
 
  • #9
Simon Bridge said:
Yes? Where are you up to in your solution?

Gah, sorry. I just fell of the map for a week, didn't I? I basically forgot to use radians in my calculation and also forgot, that the formula applies to a potential well from ##-a## to ##a##, not ##0## to ##L##.

Should I have posted my final numerical answer in this thread? I guess that would have been in good housekeeping...

The answer I got was ##\approx 0.73nm##.
 

1. What is a finite potential well?

A finite potential well is a concept in quantum mechanics that describes a region in space where a particle is confined by a potential energy barrier on both sides.

2. How does the width of a finite potential well affect a particle's behavior?

The width of a finite potential well determines the allowed energy levels and the probability of finding a particle within the well. A wider well will have a larger number of energy levels and a higher probability of finding the particle within the well.

3. What is the relationship between the width of a finite potential well and the energy levels within it?

The energy levels within a finite potential well are inversely proportional to the width of the well. This means that as the width increases, the energy levels decrease, and vice versa.

4. How does the potential energy barrier influence the width of a finite potential well?

The potential energy barrier acts as a confinement for the particle, limiting its movement and determining the width of the well. The higher the potential barrier, the narrower the well will be.

5. Can the width of a finite potential well be changed?

Yes, the width of a finite potential well can be changed by altering the strength of the potential energy barrier. The width can also be changed by changing the mass or energy of the particle confined within the well.

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